From the equation (sin2x)(sin6x)=(sinx)(sin7x), what is x?

To solve this equation, we'll change the products into
sums, using the next formula:

(sin a)*(sin b) =
[cos(a-b)-cos(a+b)]/2

Let a = 2x and b =
6x:

(sin2x)*(sin6x) =
[cos(2x-6x)-cos(2x+6x)]/2

(sin2x)*(sin6x) =
[cos(-4x)-cos(8x)]/2

Let a = x and b =
7x:

(sinx)*(sin7x) =
[cos(x-7x)-cos(x+7x)]/2

(sinx)*(sin7x) =
[cos(-6x)-cos(8x)]/2

We'll re-write the
equation:

[cos(-4x)-cos(8x)]/2 =
[cos(-6x)-cos(8x)]/2

[cos(-4x)-cos(8x)] =
[cos(-6x)-cos(8x)]

We'll eliminate
cos(8x):

cos(-4x) =
cos(-6x)

Since the cosine function is even, we'll
get:

cos (4x) = cos (6x)

We'll
subtract cos (6x) both sides:

cos (4x) - cos (6x) =
0

Since the trigonometric functions are matching, we'll
transfrom the difference into a product, usinfg the
formula:

cos a - cos b = -2sin
[(a+b)/2]*sin[(a-b)/2]

cos 4x - cos 6x = -2sin
[(4x+6x)/2]*sin[(4x-6x)/2]

cos 4x - cos 6x = 2(sin 5x)*(sin
x)

We'll re-write the
equation:

2(sin 5x)*(sin x) =
0

We'll cancel each
factor:

sin 5x = 0

5x =
(-1)^k*arcsin 0 + k*pi

5x =
k*pi

x = k*pi/5

sin x =
0

x = k*pi

The
solutions of the equation are represented by the reunion of sets: {k*pi/5 / k is integer
number}U{k*pi / k is integer number}.